A particle of mass m moves along a circle of radius R with a normal acceleration varying with time as wn = at2, where a is a constant. Find the time dependence of the power developed by all the forces acting on the particle, and the mean value of this power averaged over the first t seconds after the beginning of motion.
The power of all the forces is P = m a R t, a straight line from zero. The mean power over the first t seconds is m a R t/2. With the numbers on screen, for t = s the mean power is .
Problem statement
Problem 1.87. A particle of mass m moves along a circle of radius R with a normal acceleration varying with time as wn = at2, where a is a constant. Find the time dependence of the power developed by all the forces acting on the particle, and the mean value of this power averaged over the first t seconds after the beginning of motion.
Interpretation. A particle of mass m moves along a circle of radius R with a normal acceleration that varies with time as wn = at², where a is a constant. Find the time dependence of the power developed by all the forces acting on the particle, and the mean value of this power averaged over the first t seconds after the beginning of motion.
We assume the particle starts from rest at t = 0 and moves in one direction round the circle. The speed then grows from zero.
We assume m = kg, R = m, a = m/s⁴ and a window of t = s as starting values, since the problem gives only symbols. All four are sliders.
We assume the forces are a string or track force along the radius and a force along the path. Together they are all the forces that act, so their power is the power asked for.
Symbols: wn is the acceleration toward the centre, wt the acceleration along the path, v the speed.
The big idea in plain words
A force does work only through the part of it that lies along the motion. On a circle the pull toward the centre is always at right angles to the motion, so it does no work and gives no power. The speed changes only because of the force along the path. So the power of all the forces equals the power of that tangential force alone: P = m wt v.
The solution in each frame
Only the ground frame is needed. The centre of the circle is at rest in it, and power is defined in it.
The solution step by step
Step 1, the speed. For motion on a circle the normal acceleration is wn = v²/R. Setting v²/R = at² gives v = √(aR) t. With the numbers, v = t m/s. At the end of the window v = m/s.
Step 2, the tangential acceleration. The speed is proportional to t, so wt = dv/dt = √(aR) = m/s². It is constant.
Step 3, the power at one moment. Power is the force dotted with the velocity. The force is m(wt along the path + wn toward the centre). The velocity lies along the path, so the normal part gives zero. P = m wt v = m √(aR) × √(aR) t = m a R t. With the numbers, P = t W.
Step 4, the mean power. The mean of P over the first t seconds is the integral of P dt divided by t. The integral of m a R t dt is m a R t²/2. So the mean is m a R t/2 = for t = s.
Step 5, the cases. The mean power is half the power at the end of the window, , because P grows in a straight line from zero. In this window the particle goes round times.
A separate calculation does not use the power formula. It follows the angle round the circle in 40 000 small steps, builds the position, then finds the velocity and the acceleration by differences of position. At each step it adds F dotted with v times the small time. It gets a work of J and a mean power of W, against J and W from the formulas. The final speed is m/s against m/s. The largest relative difference is .
Five methods
1. Force method
Split the force into the part along the path and the part along the radius. Only the first does work. This is steps 1 to 3.
2. Energy method
The work of all forces equals the gain of kinetic energy, ½mv² = ½ m a R t². Dividing by t gives the mean power m a R t/2. Taking the slope of the energy with time gives P = m a R t.
3. Graph method
P against t is a straight line through zero. The area under it up to t is a triangle. Dividing that area by t gives half the height of the line, which is the mean power.
4. Calculus method
P = dK/dt with K = ½ m a R t². The mean is (1/t) ∫ P dt = m a R t/2. These are steps 3 and 4.
5. Check method
Units: m a R has units kg · (m/s⁴) · m = kg m²/s⁴. Times t that is kg m²/s³, which is watts. Limits: at t = 0 the power is zero, since the particle is at rest. Doubling m doubles every power.
Fastest: the energy method. Most reliable: the force method with the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Mass m | Yes | The power, the work and the force. Not the speed or the path |
| Radius R | Yes | The speed, the tangential acceleration, the power and the turns made |
| Constant a | Yes | The speed, both accelerations and the power |
| Time window t | Yes | How far the motion goes, the final power and the mean power |
| Given-answer mode | Sets t | The page finds the window for a chosen mean power, and warns if it is outside 1 s to 6 s |
| Play and time bar | No | Play runs the window in real time, holds 1.2 s, then repeats |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Mass | m | kg |
| Radius | R | m |
| Constant | a | m/s⁴ |
| Time window | t | s |
| Tangential acceleration | √(aR) | m/s² |
| Speed at the end | v | m/s |
| Normal acceleration at the end | wn | m/s² |
| Power at the end | P | |
| Mean power |
Exam traps
- Adding the power of the normal force. It is at right angles to v, so its power is zero.
- Writing the mean power as the power at time t. The mean is half of it.
- Taking the tangential acceleration as zero because the speed on a circle looks steady. Here v grows, so wt is √(aR).
- Using wn = at² as the tangential acceleration. It is the acceleration toward the centre.
- Forgetting that the particle starts from rest, so v = 0 at t = 0.